If S is the set of distinct values of 'b' for which the following system of linear equations
`x+y+z=1`
`x+ay +z=1`
`ax+by+z=0`
has no solution then S is

Options

A. an infinite set
B. a finite set containing two or more elements
C. singleton set
D. an empty set

Answer

C

Explanation

` because triangle =|(1,1,1),(1,a,1),(a,b,1)|=1(a-b)-1(1-a)+1(b-a^(2))`
`= -(a-1)^(2)`
` triangle_(1) =|(1,1,1),(1,a,1),(0,b,1)|=1(a-b)-1(1)+1(b)= -(a-1)`
` triangle_(2) =|(1,1,1),(1,1,1),(a,0,1)|=1(1)-1(1-a)+1(0-a)=0`
`and triangle_(3) =|(1,1,1),(1,a,1),(a,b,0)|=1(-b)-1(-a)+1(b-a^(2))`
`=-a(a-1)`
For `a=1`
`triangle =triangle_(1)=triangle_(2)=triangle_(3) =0`
`triangle " for " b=1` only
`x+y+z=1`,
`x+y+z=1`
`and x+y+z=0`
i.e. no solution ` " " `( therefore RHS is not equal)
Hence, for no solution b=1 only